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3 years ago

In the figure, if AB || CD, then angle EFD is equal to:(a) 50(d) 40(c) 20(d) 30

Mathematics

2 answers:

dolphi86 [110]3 years ago

8 0

Answer:

d. 40°

Step-by-step explanation:

70-30=40°

Evgesh-ka [11]3 years ago

4 0

Answer:

d) 40

Step-by-step explanation:

AB || CD, so ∠DCB + ∠ABC = 180 { co interior angles }

∠DCB + 70 = 180

∠DCB = 180 -70 = 110

∠FCE = ∠DCB { Vertically opposite angles }

∠FCE = 110

In triangle CFE

∠EFD + ∠CEF + ∠FCE = 180 ( sum of angles of triangle)

∠EFD + 30 + 110 = 180

∠EFD + 140 = 180

∠EFD = 180 -140

∠EFD = 40

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A hexagonal prism has a height of 6.5 centimeters and a base with an area of 30.5

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Answer:

198.25cm³

Solution

V=ABh=30.5·6.5=198.25cm³

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3 years ago

how many 5 letter combinations can be created from the letters in the world “friendly” assume that the order of the letters does

Artyom0805 [142]

Answer:

Number of combinations = 56

Step-by-step explanation:

Points to remember

nCₐ = n!/(r!(n - r)!)

The given word is "FRIENDLY"

F R I E N D L Y is a 8 letter word.

<u>To find the number of combinations</u>

5 letters can be selected from 8 letter = 8C₅

8C₅ = 8!/(8 - 5)!5!

= 8!/3!5!

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= 8 * 7

= 56

5 letter combinations can be created from the letters in the world “friendly = 56

4 0

3 years ago

A coach is assessing the correlation between the number of hours spent practicing and the average number of points scored in a g

cricket20 [7]

Answer:

a) $r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1$

We have a perfect linear relationship between the two variables

b) $m=\frac{90}{15}=6$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2$

$\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17$

And we can find the intercept using this:

$b=\bar y -m \bar x=17-(6*2)=5$

So the line would be given by:

$y=6 x +5$

c) For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

Step-by-step explanation:

We have the following data:

Number of hours spent practicing (x) 0 0.5 1 1.5 2 2.5 3 3.5 4

Score in the game (y) 5 8 11 14 17 20 23 26 29

Part a

The correlation coefficient is given:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

For our case we have this:

n=9 $\sum x = 18, \sum y = 153, \sum xy = 396, \sum x^2 =51, \sum y^2 =3141$

$r=\frac{9(396)-(18)(153)}{\sqrt{[9(51) -(18)^2][9(3141) -(153)^2]}}=1$

We have a perfect linear relationship between the two variables

Part b

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=51-\frac{18^2}{9}=15$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=396-\frac{18*153}{9}=90$

And the slope would be:

$m=\frac{90}{15}=6$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{18}{9}=2$

$\bar y= \frac{\sum y_i}{n}=\frac{153}{9}=17$

And we can find the intercept using this:

$b=\bar y -m \bar x=17-(6*2)=5$

So the line would be given by:

$y=6 x +5$

Part c

For this case the slope indicates that for each increase of the number of hours in 1 unit we have an expected increase in the score about 6 units.

And the intercept 5 represent the minimum score expected for any game

5 0

3 years ago

How do i simplify -9x^3+8x^2-(7x/6)-(-7)

cricket20 [7]

Multiply -1 by -7
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3 years ago

If you had 36 cookies and gave 1/2 to one friend and 1/4 to another, how many could you have left?

Ann [662]

Step-by-step explanation:

$\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$

$\frac{3}{4} 36 = 27$

6 0

2 years ago

In the figure, if AB || CD, then angle EFD is equal to:(a) 50(d) 40(c) 20(d) 30​

In the figure, if AB || CD, then angle EFD is equal to:(a) 50(d) 40(c) 20(d) 30